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12-24 Chapter 12 
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 12.75. IDENTIFY: Apply Newton’s second law to the ingot. Use the expression for the buoyancy force given by 
Archimedes’s principle to solve for the volume of the ingot. Then use the facts that the total mass is the 
mass of the gold plus the mass of the aluminum and that the volume of the ingot is the volume of the gold 
plus the volume of the aluminum. 
SET UP: The free-body diagram for the piece of alloy is given in Figure 12.75. 
 
 EXECUTE: y yF ma=∑ 
tot 0B T m g+ − = 
totB m g T= − 
45 0 N 39 0 N 6 0 NB = . − . = . 
Figure 12.75 
 
Also, tot 45 0 Nm g = . so 
2
tot 45 0 N/(9 80 m/s ) 4 59 kg.m = . . = . 
We can use the known value of the buoyant force to calculate the volume of the object: 
w obj 6 0 NB V gρ= = . 
4 3
obj 3 2
w
6.0 N 6.0 N 6.122 10 m
(1000 kg/m )(9.80 m/s )
V
gρ
−= = = × 
We know two things: 
(1) The mass gm of the gold plus the mass am of the aluminum must add to tot:m g a totm m m+ = 
We write this in terms of the volumes gV and aV of the gold and aluminum: g g a a totV V mρ ρ+ = 
(2) The volumes aV and gV must add to give obj:V a g objV V V+ = so that a obj gV V V= − 
Use this in the equation in (1) to eliminate a:V g g a obj g tot( )V V V mρ ρ+ − = 
3 3 4 3
tot a obj 4 3
g 3 3 3 3
g a
4.59 kg (2.7 10 kg/m )(6.122 10 m ) 1.769 10 m .
19.3 10 kg/m 2.7 10 kg/m
m V
V
ρ
ρ ρ
−
−− − × ×= = = ×
− × − ×
 
Then 3 3 4 3g g g (19.3 10 kg/m )(1.769 10 m ) 3.41 kgm Vρ
−= = × × = and the weight of gold is 
g g 33 4 N.w m g= = . 
EVALUATE: The gold is 29% of the volume but 74% of the mass, since the density of gold is much 
greater than the density of aluminum. 
 12.76. IDENTIFY: Apply y yF ma=∑ to the ball, with y+ upward. The buoyant force is given by Archimedes’s 
principle. 
SET UP: The ball’s volume is 3 3 34 4 (12 0 cm) 7238 cm .
3 3
V rπ π= = . = As it floats, it displaces a weight of 
water equal to its weight. 
EXECUTE: (a) By pushing the ball under water, you displace an additional amount of water equal to 
76.0% of the ball’s volume or 3 3(0 760)(7238 cm ) 5501cm .. = This much water has a mass of 
5501 5 501kgg = . and weighs 2(5 501 kg)(9 80 m/s ) 53 9 N,. . = . which is how hard you’ll have to push to 
submerge the ball. 
(b) The upward force on the ball in excess of its own weight was found in part (a): 53.9 N. The ball’s mass 
is equal to the mass of water displaced when the ball is floating: 
3 3(0 240)(7238 cm )(1 00 g/cm ) 1737 g 1 737 kg,. . = = . 
and its acceleration upon release is thus 2net 53.9 N 31.0 m/s .
1.737 kg
Fa
m
= = = 
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 Fluid Mechanics 12-25 
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EVALUATE: When the ball is totally immersed the upward buoyant force on it is much larger than its 
weight. 
 12.77. (a) IDENTIFY: Apply Newton’s second law to the crown. The buoyancy force is given by Archimedes’s 
principle. The target variable is the ratio c w/ (c crown, w water).ρ ρ = = 
SET UP: The free-body diagram for the crown is given in Figure 12.77. 
 
 EXECUTE: y yF ma=∑ 
0T B w+ − = 
T fw= 
w c ,B V gρ= where w densityρ = 
of water, c volumeV = of crown 
Figure 12.77 
 
Then w c 0.fw V g wρ+ − = 
w c(1 )f w V gρ− = 
Use c c ,w V gρ= where c densityρ = of crown. 
c c w c(1 )f V g V gρ ρ− = 
c
w
1 ,
1 f
ρ
ρ
=
−
 as was to be shown. 
0f → gives c w/ 1ρ ρ = and 0.T = These values are consistent. If the density of the crown equals the 
density of the water, the crown just floats, fully submerged, and the tension should be zero. 
When 1,f → c wρ ρ>> and .T w= If c wρ ρ>> then B is negligible relative to the weight w of the 
crown and T should equal w. 
(b) “apparent weight” equals T in the rope when the crown is immersed in water. ,T fw= so need to 
compute f. 
3 3
c 19 3 10 kg/m ;ρ = . × 
3 3
w 1 00 10 kg/mρ = . × 
c
w
1
1 f
ρ
ρ
=
−
 gives 
3 3
3 3
19 3 10 kg/m 1
11 00 10 kg/m f
. × =
−. ×
 
19 3 1/(1 )f. = − and 0 9482f = . 
Then (0 9482)(12 9 N) 12 2 N.T fw= = . . = . 
(c) Now the density of the crown is very nearly the density of lead; 
3 3
c 11 3 10 kg/m .ρ = . × 
c
w
1
1 f
ρ
ρ
=
−
 gives 
3 3
3 3
11 3 10 kg/m 1
11 00 10 kg/m f
. × =
−. ×
 
11 3 1/(1 )f. = − and 0 9115f = . 
Then (0 9115)(12 9 N) 11 8 N.T fw= = . . = . 
EVALUATE: In part (c) the average density of the crown is less than in part (b), so the volume is greater. 
B is greater and T is less. These measurements can be used to determine if the crown is solid gold, without 
damaging the crown. 
 12.78. IDENTIFY: Problem 12.77 says object
fluid
1 ,
1 f
ρ
ρ
=
−
 where the apparent weight of the object when it is totally 
immersed in the fluid is fw. 
SET UP: For the object in water, water water /f w w= and for the object in the unknown fluid, 
fluid fluid / .f w w= 
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12-26 Chapter 12 
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EXECUTE: (a) steel
fluid fluid
,w
w w
ρ
ρ
=
−
 steel
water water
.w
w w
ρ
ρ
=
−
 Dividing the second of these by the first gives 
fluid fluid
water water
.w w
w w
ρ
ρ
−=
−
 
(b) When fluidw is greater than water ,w the term on the right in the above expression is less than one, 
indicating that the fluid is less dense than water, and this is consistent with the buoyant force when 
suspended in liquid being less than that when suspended in water. If the density of the fluid is the same as 
that of water fluid water ,w w= as expected. Similarly, if fluidw is less than water ,w the term on the right in 
the above expression is greater than one, indicating that the fluid is more dense than water. 
(c) Writing the result of part (a) as fluid fluid
water water
1 ,
1
f
f
ρ
ρ
−=
−
 and solving for fluid ,f 
fluid
fluid water
water
1 (1 ) 1 (1.220)(0.128) 0.844 84.4%.f fρ
ρ
= − − = − = = 
EVALUATE: Formic acid has density greater than the density of water. When the object is immersed in 
formic acid the buoyant force is greater and the apparent weight is less than when the object is immersed in 
water. 
 12.79. IDENTIFY and SET UP: Use Archimedes’s principle for B. 
(a) water tot ,B V gρ= where totV is the total volume of the object. 
tot m 0,V V V= + where mV is the volume of the metal. 
EXECUTE: m m/V w gρ= so tot m 0/V w g Vρ= + 
This gives water m 0( / ).B g w g Vρ ρ= + 
Solving for 0V gives 0 water m/( ) /( ),V B g w gρ ρ= − as was to be shown. 
(b) The expression derived in part (a) gives 
4 3
0 3 2 3 3 2
20 N 156 N 2.52 10 m
(1000 kg/m )(9.80 m/s ) (8.9 10 kg/m )(9.80 m/s )
V −= − = ×
×
 
3 3
tot 3 2
water
20 N 2.04 10 m
(1000 kg/m )(9.80 m/s )
BV
gρ
−= = = × and 
4 3 3 3
0 tot/ (2.52 10 m )/(2.04 10 m ) 0.124.V V
− −= × × = 
EVALUATE: When 0 0,V → the object is solid and obj m m/( ).V V w gρ= = For 0 0,V = the result in part (a) 
gives m water m water obj water( / ) ,B w V g V gρ ρ ρ ρ= = = which agrees with Archimedes’s principle. As 0V 
increases with the weight kept fixed, the total volume of the object increases and there is an increase in B. 
 12.80. IDENTIFY: For a floating object the buoyant force equals the weight of the object.Archimedes’s principle 
says the buoyant force equals the weight of fluid displaced by the object. .m Vρ= 
SET UP: Let d be the depth of the oil layer, h the depth that the cube is submerged in the water and L be 
the length of a side of the cube. 
EXECUTE: (a) Setting the buoyant force equal to the weight and canceling the common factors of g and 
the cross-sectional area, (1000) (750) (550) .h d L+ = d, h and L are related by 0 35 ,d h L L+ + . = so 
0 65 .h L d= . − Substitution into the first relation gives (0 65)(1000) (550) 2 0 040 m.(1000) (750) 5 00
Ld L . −= = = . − . 
(b) The gauge pressure at the lower face must be sufficient to support the block (the oil exerts only 
sideways forces directly on the block), and 3 2wood (550 kg/m )(9 80 m/s )(0 100 m) 539 Pa.p gLρ= = . . = 
EVALUATE: As a check, the gauge pressure, found from the depths and densities of the fluids, is 
3 3 2[(0 040 m)(750 kg/m ) (0 025 m)(1000 kg/m )](9 80 m/s ) 539 Pa.. + . . = 
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 Fluid Mechanics 12-27 
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 12.81. IDENTIFY and SET UP: Apply the first condition of equilibrium to the barge plus the anchor. Use 
Archimedes’s principle to relate the weight of the boat and anchor to the amount of water displaced. In 
both cases the total buoyant force must equal the weight of the barge plus the weight of the anchor. Thus 
the total amount of water displaced must be the same when the anchor is in the boat as when it is over the 
side. When the anchor is in the water the barge displaces less water, less by the amount the anchor 
displaces. Thus the barge rises in the water. 
EXECUTE: The volume of the anchor is 3 3 3anchor / (35.0 kg)/(7860 kg/m ) 4.453 10 m .V m ρ
−= = = × The 
barge rises in the water a vertical distance h given by 3 34.453 10 m ,hA −= × where A is the area of the 
bottom of the barge. 3 3 2 4(4.453 10 m )/(8.00 m ) 5.57 10 m.h − −= × = × 
EVALUATE: The barge rises a very small amount. The buoyancy force on the barge plus the buoyancy 
force on the anchor must equal the weight of the barge plus the weight of the anchor. When the anchor is in 
the water, the buoyancy force on it is less than its weight (the anchor doesn’t float on its own), so part of the 
buoyancy force on the barge is used to help support the anchor. If the rope is cut, the buoyancy force on the 
barge must equal only the weight of the barge and the barge rises still farther. 
 12.82. IDENTIFY: Apply y yF ma=∑ to the barrel, with y+ upward. The buoyant force on the barrel is given by 
Archimedes’s principle. 
SET UP: av tot / .m Vρ = An object floats in a fluid if its average density is less than the density of the fluid. 
The density of seawater is 31030 kg/m . 
EXECUTE: (a) The average density of a filled barrel is 
3 3oil steel steel
oil 3
15 0 kg750 kg/m 875 kg/m ,
0 120 m
mm m
VV
ρ+ .= + = + =
.
 which is less than the density of 
seawater, so the barrel floats. 
(b) The fraction above the surface (see Problem 12.29) is 
3
av
3
water
875 kg/m1 1 0 150 15 0%.
1030 kg/m
ρ
ρ
− = − = . = . 
(c) The average density is 3 33
32 0 kg910 kg/m 1172 kg/m ,
0 120 m
. + =
.
 which means the barrel sinks. In order to 
lift it, a tension 
3 3 2 3 3 2
tot (1177 kg/m )(0 120 m )(9 80 m/s ) (1030 kg/m )(0 120 m )(9 80 m/s ) 173 NT w B= − = . . − . . = is 
required. 
EVALUATE: When the barrel floats, the buoyant force B equals its weight, w. In part (c) the buoyant force 
is less than the weight and .T w B= − 
 12.83. IDENTIFY: Apply Newton’s second law to the block. In part (a), use Archimedes’s principle for the 
buoyancy force. In part (b), use Eq. (12.6) to find the pressure at the lower face of the block and then use 
Eq. (12.3) to calculate the force the fluid exerts. 
(a) SET UP: The free-body diagram for the block is given in Figure 12.83a. 
 
 EXECUTE: y yF ma=∑ 
0B mg− = 
L sub B objV g V gρ ρ= 
Figure 12.83a 
 
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The fraction of the volume that is submerged is sub obj B L/ / .V V ρ ρ= 
Thus the fraction that is above the surface is above obj B L/ 1 / .V V ρ ρ= − 
EVALUATE: If B Lρ ρ= the block is totally submerged as it floats. 
(b) SET UP: Let the water layer have depth d, as shown in Figure 12.83b. 
 
 
EXECUTE: 0 w L ( )p p gd g L dρ ρ= + + − 
Applying y yF ma=∑ to the block gives 
0( ) 0.p p A mg− − = 
Figure 12.83b 
 
w L B[ ( )]gd g L d A LAgρ ρ ρ+ − = 
A and g divide out and w L B( )d L d Lρ ρ ρ+ − = 
w L B L( ) ( )d Lρ ρ ρ ρ− = − 
L B
L w
d Lρ ρ
ρ ρ
⎛ ⎞−= ⎜ ⎟−⎝ ⎠
 
(c) 
3 3 3 3
3 3 3
13 6 10 kg/m 7 8 10 kg/m (0 100 m) 0 0460 m 4 60 cm
13 6 10 kg/m 1000 kg/m
d
⎛ ⎞. × − . ×= . = . = .⎜ ⎟⎜ ⎟. × −⎝ ⎠
 
EVALUATE: In the expression derived in part (b), if B Lρ ρ= the block floats in the liquid totally 
submerged and no water needs to be added. If L wρ ρ→ the block continues to float with a fraction 
B w1 /ρ ρ− above the water as water is added, and the water never reaches the top of the block ( ).d → ∞ 
 12.84. IDENTIFY: For the floating tanker, the buoyant force equals its total weight. The buoyant force is given by 
Archimedes’s principle. 
SET UP: When the metal is in the tanker, it displaces its weight of water and after it has been pushed 
overboard it displaces its volume of water. 
EXECUTE: (a) The change in height yΔ is related to the displaced volume by ,VV y
A
ΔΔ Δ = where A is 
the surface area of the water in the lock. VΔ is the volume of water that has the same weight as the metal, 
so 
6
water
3 3 2
water
/( ) (2 50 10 N) 0 213 m.
(1 00 10 kg/m )(9 80 m/s )[(60 0 m)(20 0 m)]
V w g wy
A A gA
ρ
ρ
Δ . ×Δ = = = = = .
. × . . .
 
(b) In this case, VΔ is the volume of the metal; in the above expression, waterρ is replaced by 
metal water9 00 ,ρ ρ= . which gives 
8, and 0 189 m;
9 9
yy y y yΔΔ ′ = Δ − Δ ′ = Δ = . the water level falls this 
amount. 
EVALUATE: The density of the metal is greater than the density of water, so the volume of water that has 
the same weight as the steel is greater than the volume of water that has the same volume as the steel. 
 12.85. IDENTIFY: Consider the fluid in the horizontal part of the tube. This fluid, with mass ,Alρ is subject to a 
net force due to the pressure difference between the ends of the tube. 
SET UP: The difference between the gauge pressures at the bottoms of the ends of the tubes is 
L R( ).g y yρ − 
EXECUTE: The net force on the horizontal part of the fluid is L R( ) ,g y y A Alaρ ρ− = or, L R( ) .
ay y l
g
− = 
(b) Again consider the fluid in the horizontal part of the tube. As in part (a), the fluid is accelerating; the 
center of mass has a radial acceleration of magnitude 2rad /2,a lω= and so the difference in heights 
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 Fluid Mechanics 12-29 
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between the columns is 2 2 2( /2)( / ) /2 .l l g l gω ω= An equivalent way to do part (b) is to break the fluid in 
the horizontal part of the tube into elements of thickness dr; the pressure difference between the sides of 
this piece is2( )dp r drρ ω= and integrating from 2 20 to gives /2,r r l p lρω= = Δ = the same result. 
EVALUATE: (c) The pressure at the bottom of each arm is proportional to ρ and the mass of fluid in the 
horizontal portion of the tube is proportional to ,ρ so ρ divides out and the results are independent of the 
density of the fluid. The pressure at the bottom of a vertical arm is independent of the cross-sectional area 
of the arm. Newton’s second law could be applied to a cross-sectional of fluid smaller than that of the 
tubes. Therefore, the results are independent and of the size and shape of all parts of the tube. 
 12.86. IDENTIFY: Apply m=∑F a to a small fluid element located a distance r from the axis. 
SET UP: For rotational motion, 2 .a rω= 
EXECUTE: (a) The change in pressure with respect to the vertical distance supplies the force necessary to 
keep a fluid element in vertical equilibrium (opposing the weight). For the rotating fluid, the change in 
pressure with respect to radius supplies the force necessary to keep a fluid element accelerating toward the 
axis; specifically, ,pdp dr a drr ρ
∂= = ∂ and using 
2a rω= gives 2 .p rr ρω
∂ =∂ 
(b) Let the pressure at 0, 0y r= = be ap (atmospheric pressure); integrating the expression for 
p
r
∂
∂
 from 
part (a) gives 
2
2
a( , 0) .2
p r y p r
ρω
 = = + 
(c) In Eq. (12.5), 2 a 1, ( , 0)p p p p p r y= = = = as found in part (b), 1 20 and ( ),y y h r= = the height of the 
liquid above the 0y = plane. Using the result of part (b) gives 2 2( ) /2 .h r r gω= 
EVALUATE: The curvature of the surface increases as the speed of rotation increases. 
 12.87. IDENTIFY: Follow the procedure specified in part (a) and integrate this result for part (b). 
SET UP: A rotating particle a distance r′ from the rotation axis has inward acceleration 2 .rω ′ 
EXECUTE: (a) The net inward force is ( ) ,p dp A pA Adp+ − = and the mass of the fluid element is 
.Adrρ ′ Using Newton’s second law, with the inward radial acceleration of 2 ,rω ′ gives 2 .dp r drρω= ′ ′ 
(b) Integrating the above expression, 2
0 0
p r
p r
dp r drρω= ′ ′∫ ∫ and 
2
2 2
0 0( ),2
p p r rρω
⎛ ⎞
− = −⎜ ⎟⎜ ⎟
⎝ ⎠
 which is the 
desired result. 
(c) The net force on the object must be the same as that on a fluid element of the same shape. Such a fluid 
element is accelerating inward with an acceleration of magnitude 2 cm,Rω and so the force on the object is 
2
cm.V Rρ ω 
(d) If cm ob cm ob,R Rρ ρ> the inward force is greater than that needed to keep the object moving in a circle 
with radius cm obR at angular frequency ,ω and the object moves inward. If cm ob cm ob,R Rρ ρ< the net 
force is insufficient to keep the object in the circular motion at that radius, and the object moves outward. 
(e) Objects with lower densities will tend to move toward the center, and objects with higher densities will 
tend to move away from the center. 
EVALUATE: The pressure in the fluid increases as the distance r from the rotation axis increases. 
 12.88. IDENTIFY: Follow the procedure specified in the problem. 
SET UP: Let increasing x correspond to moving toward the back of the car. 
EXECUTE: (a) The mass of air in the volume element is ,dV Adxρ ρ= and the net force on the element in 
the forward direction is ( ) .p dp A pA Adp+ − = From Newton’s second law, ( ) ,Adp Adx aρ= from which 
.dp adxρ= 
(b) With ρ given to be constant, and with 0p p= at 0,x = 0 .p p axρ = + 
(c) Using 31.2 kg/mρ = in the result of part (b) gives 
3 2 5
atm(1.2 kg/m )(5.0 m/s )(2.5 m) 15.0 Pa 15 10 ,p
−= = × so the fractional pressure difference is negligible. 
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(d) Following the argument in Section 12.3, the force on the balloon must be the same as the force on the 
same volume of air; this force is the product of the mass Vρ and the acceleration, or .Vaρ 
(e) The acceleration of the balloon is the force found in part (d) divided by the mass bal bal, or ( / ) .V aρ ρ ρ 
The acceleration relative to the car is the difference between this acceleration and the car’s acceleration, 
rel bal[( / ) 1] .a aρ ρ= − 
(f) For a balloon filled with air, bal( / ) 1ρ ρ < (air balloons tend to sink in still air), and so the quantity in 
square brackets in the result of part (e) is negative; the balloon moves to the back of the car. For a helium 
balloon, the quantity in square brackets is positive, and the balloon moves to the front of the car. 
EVALUATE: The pressure in the air inside the car increases with distance from the windshield toward the 
rear of the car. This pressure increase is proportional to the acceleration of the car. 
 12.89. IDENTIFY: After leaving the tank, the water is in free fall, with 0xa = and .ya g= + 
SET UP: From Example 12.8, the speed of efflux is 2 .gh 
EXECUTE: (a) The time it takes any portion of the water to reach the ground is 2( ) ,H ht
g
−= in which 
time the water travels a horizontal distance 2 ( ).R vt h H h= = − 
(b) Note that if , ( ) ( ) ,h H h h H h H h h′ = − ′ − ′ = − and so h H h′ = − gives the same range. A hole H h− 
below the water surface is a distance h above the bottom of the tank. 
EVALUATE: For the special case of /2, h H h h= = ′ and the two points coincide. For the upper hole the 
speed of efflux is less but the time in the air during the free fall is greater. 
12.90. IDENTIFY: Use Bernoulli’s equation to find the velocity with which the water flows out the hole. 
SET UP: The water level in the vessel will rise until the volume flow rate into the vessel, 4 32.40 10 m /s,−× 
equals the volume flow rate out the hole in the bottom. 
 
 Let points 1 and 2 be chosen as in 
Figure 12.90. 
Figure 12.90 
 
EXECUTE: Bernoulli’s equation: 2 21 11 1 1 2 2 22 2p gy v p gy vρ ρ ρ ρ+ + = + + 
Volume flow rate out of hole equals volume flow rate from tube gives that 4 32 2 2.40 10 m /sv A
−= × and 
4 3
2 4 2
2.40 10 m /s 1.60 m/s
1.50 10 m
v
−
−
×= =
×
 
1 2A A and 1 1 2 2v A v A= says that 
2 21 1
1 22 2 ;v vρ ρ neglect the 
21
12 vρ term. 
Measure y from the bottom of the bucket, so 2 0y = and 1y h= . 
1 2 ap p p= = (air pressure) 
Then 21a a 22p gh p vρ ρ+ = + and 
2 2 2
2 /2 (1.60 m/s) /2(9.80 m/s ) 0.131 m 13.1 cmh v g= = = = 
EVALUATE: The greater the flow rate into the bucket, the larger 2v will be at equilibrium and the higher 
the water will rise in the bucket. 
12.91. IDENTIFY: Apply Bernoulli’s equation and the equation of continuity. 
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SET UP: Example 12.8 says the speed of efflux is 2 ,gh where h is the distance of the hole below the 
surface of the fluid. 
EXECUTE: (a) 2 2 33 3 1 3 32 ( ) 2(9.80 m/s )(8.00 m)(0.0160 m ) 0.200 m /s.v A g y y A= − = = 
(b) Since 3p is atmospheric pressure, the gauge pressure at point 2 is 
2
2 2 2 3
2 3 2 3 1 3
2
1 1 8( ) 1 ( ),
2 2 9
Ap v v v g y y
A
ρ ρ ρ
⎛ ⎞⎛ ⎞⎜ ⎟= − = − = −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
 using the expression for 3v found above. 
Substitution of numerical values gives 42 6.97 10 Pa.p = × 
EVALUATE: We could also calculate 2p by applying Bernoulli’s equation to points 1 and 2. 
12.92. IDENTIFY: Apply Bernoulli’s equation to the air in the hurricane. 
SET UP: For a particle a distance r from the axis, the angular momentum is .L mvr= 
EXECUTE: (a) Using theconstancy of angular momentum, the product of the radius and speed is constant, 
so the speed at the rim is about 30(200 km/h) 17 km/h.
350
⎛ ⎞ =⎜ ⎟
⎝ ⎠
 
(b) The pressure is lower at the eye, by an amount 
2
3 2 2 31 1 m/s(1.2 kg/m )((200 km/h) 17 km/h ) 1.8 10 Pa.
2 3.6 km/h
p ⎛ ⎞Δ = − ( ) = ×⎜ ⎟
⎝ ⎠
 
(c) 
2
160 m.
2
v
g
= 
(d) The pressure difference at higher altitudes is even greater. 
EVALUATE: According to Bernoulli’s equation, the pressure decreases when the fluid velocity increases. 
 12.93. IDENTIFY: Apply Bernoulli’s equation and the equation of continuity. 
SET UP: Example 12.8 shows that the speed of efflux at point D is 12 .gh 
EXECUTE: Applying the equation of continuity to points at C and D gives that the fluid speed is 18gh at 
C. Applying Bernoulli’s equation to points A and C gives that the gauge pressure at C is 
1 1 14 3 ,gh gh ghρ ρ ρ− = − and this is the gauge pressure at the surface of the fluid at E. The height of the 
fluid in the column is 2 13 .h h= 
EVALUATE: The gauge pressure at C is less than the gauge pressure 1ghρ at the bottom of tank A because 
of the speed of the fluid at C. 
 12.94. IDENTIFY: Apply Bernoulli’s equation to points 1 and 2. Apply 0p p ghρ= + to both arms of the U-shaped 
tube in order to calculate h. 
SET UP: The discharge rate is 1 1 2 2.v A v A= The density of mercury is 
3 3
m 13.6 10 kg/mρ = × and the 
density of water is 3 3w 1.00 10 kg/m .ρ = × Let point 1 be where 
4 2
1 40.0 10 mA
−= × and point 2 is where 
4 2
2 10.0 10 m .A
−= × 1 2.y y= 
EXECUTE: (a) 
3 3
1 4 2
6.00 10 m /s 1.50 m/s.
40.0 10 m
v
−
−
×= =
×
 
3 3
2 4 2
6.00 10 m /s 6.00 m/s
10.0 10 m
v
−
−
×= =
×
 
(b) 2 21 11 1 1 2 2 22 2 .p gy v p gy vρ ρ ρ ρ+ + = + + 
2 2 3 2 2 41 1
1 2 2 12 2( ) (1000 kg/m )([6.00 m/s] [1.50 m/s] ) 1.69 10 Pap p v vρ− = − = − = × 
(c) 1 w 2 mp gh p ghρ ρ+ = + and 
4
1 2
3 3 3 3 2
m w
1.69 10 Pa 0.137 m 13.7 cm.
( ) (13.6 10 kg/m 1.00 10 kg/m )(9.80 m/s )
p ph
gρ ρ
− ×= = = =
− × − ×
 
EVALUATE: The pressure in the fluid decreases when the speed of the fluid increases. 
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 12.95. (a) IDENTIFY: Apply constant acceleration equations to the falling liquid to find its speed as a function of 
the distance below the outlet. Then apply Eq. (12.10) to relate the speed to the radius of the stream. 
SET UP: 
 
 Let point 1 be at the end of the pipe 
and let point 2 be in the stream of 
liquid at a distance 2y below the 
end of the tube, as shown in 
Figure 12.95. 
Figure 12.95 
 
Consider the free fall of the liquid. Take y+ to be downward. 
Free fall implies . y ya g v= is positive, so replace it by the speed v. 
EXECUTE: 2 22 1 02 ( )v v a y y= + − gives 
2 2
2 1 22v v gy= + and 
2
2 1 22 .v v gy= + 
Equation of continuity says 1 1 2 2v A v A= 
And since 2A rπ= this becomes 2 21 1 2 2v r v rπ π= and 
2
2 1 1 2( / ) .v v r r= 
Use this in the above to eliminate 2 2 22 1 1 2 1 2: ( / ) 2v v r r v gy= + 
2 1/4
2 1 1 1 2/( 2 )r r v v gy= + 
To correspond to the notation in the problem, let 1 0v v= and 1 0,r r= since point 1 is where the liquid first 
leaves the pipe, and let 2r be r and 2y be y. The equation we have derived then becomes 
2 1/4
0 0 0/( 2 )r r v v gy= + 
(b) 0 1.20 m/sv = 
We want the value of y that gives 1 02 ,r r= or 0 2 .r r= 
The result obtained in part (a) says 4 2 4 20 0 0( 2 ) .r v gy r v+ = 
Solving for y gives 
4 2 2
0 0
2
[( / ) 1] (16 1)(1 20 m/s)
1.10 m.
2 2(9.80 m/s )
r r v
y
g
− − .
= = = 
EVALUATE: The equation derived in part (a) says that r decreases with distance below the end of the pipe. 
 12.96. IDENTIFY: Apply y yF ma∑ = to the rock. 
SET UP: In the accelerated frame, all of the quantities that depend on g (weights, buoyant forces, gauge 
pressures and hence tensions) may be replaced by ,g g a′ = + with the positive direction taken upward. 
EXECUTE: (a) The volume V of the rock is 
2
4 3
3 3 2
water water
((3.00 kg)(9.80 m/s ) 21.0 N) 8.57 10 m .
(1.00 10 kg/m )(9.80 m/s )
B w TV
g gρ ρ
−− −= = = = ×
×
 
(b) The tension is 0( ) ,
gT mg B m V g T
g
ρ ′= ′ − ′ = − ′ = where 0 21.0N.T = .g g a′ = + For 
22.50 m/s ,a = 
9.80 2.50(21.0 N) 26.4 N.
9.80
T += = 
(c) For 2 9.80 2.502.50 m/s , (21.0 N) 15.6 N.
9.80
a T −= − = = 
(d) If ,a g= − 0g ′ = and 0.T = 
EVALUATE: The acceleration of the water alters the buoyant force it exerts. 
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12.97. IDENTIFY: The sum of the vertical forces on the object must be zero. 
SET UP: The depth of the bottom of the styrofoam is not given; let this depth be 0.h Denote the length of 
the piece of foam by L and the length of the two sides by l. The volume of the object is 212 .l L 
EXECUTE: (a) The tension in the cord plus the weight must be equal to the buoyant force, so 
2 2 3 31
water foam 2( ) (0 20 m) (0 50 m)(9 80 m/s )(1000 kg/m 180 kg/m ) 80.4 N.T Vg ρ ρ= − = . . . − = 
(b) The pressure force on the bottom of the foam is ( )0 0( ) 2p gh L lρ+ and is directed up. The pressure 
on each side is not constant; the force can be found by integrating, or using the results of Problem 12.53 or 
Problem 12.55. Although these problems found forces on vertical surfaces, the result that the force is the 
product of the average pressure and the area is valid. The average pressure is 0 0( ( /(2 2))),p g h lρ+ − and 
the force on one side has magnitude 0 0( ( /(2 2)))p g h l Llρ+ − and is directed perpendicular to the side, at 
an angle of 45.0° from the vertical. The force on the other side has the same magnitude, but has a 
horizontal component that is opposite that of the other side. The horizontal component of the net buoyant 
force is zero, and the vertical component is 
2
0 0 0 0( ) 2 2(cos45.0 )( ( /(2 2))) ,2
Ll
B p gh Ll p g h l Ll gρ ρ ρ= + − ° + − = the weight of the water displaced. 
EVALUATE: The density of the object is less than the density of water, so if the cord were cut the object 
would float. When the object is fully submerged, the upward buoyant force is greater than its weight and 
the cord must pull downward on the object to hold it beneath the surface. 
 12.98. IDENTIFY: Apply Bernoulli’s equation to the fluid in the siphon. 
SET UP: Example 12.8 shows that the efflux speed from a small hole a distance h below the surface of 
fluid in a large open tank is 2 .gh 
EXECUTE: (a) The fact that the water first moves upward before leaving the siphon does not change the 
efflux speed, 2 .gh 
(b) Water will not flow if the absolute (not gauge) pressure would be negative. The hose is open to the 
atmosphere at the bottom, so the pressure at the top of the siphon is a ( ),p g H hρ− + where the 
assumption that the cross-sectional area is constant has been used to equate the speed of the liquid at the 
top and bottom. Setting 0p = and solving for H gives a( / ) .H p g hρ= − 
EVALUATE: The analysis shows that a ,pH h
gρ
+ < so there is also a limitation on .H h+ For water and 
normal atmospheric pressure, a 10.3 m.p
gρ
= 
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13-1 
 13.1. IDENTIFY and SET UP: Use the law of gravitation, Eq. (13.1), to determine g.F 
EXECUTE: S MS on M 2
SM
(S sun, M moon);
m m
F G
r
= = = E ME on M 2
EM
(E earth)m mF G
r
= = 
22
EMS on M S M S EM
2
E on M E M E SMSM
rF m m m rG
F Gm m m rr
⎛ ⎞⎛ ⎞ ⎛ ⎞
⎜ ⎟⎜ ⎟= = ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠
 
EM,r the radius of the moon’s orbit around the earth is given in Appendix F as 
83.84 10 m.× The moon is 
much closer to the earth than it is to the sun, so take the distance SMr of the moon from the sun to be SE,r 
the radius of the earth’s orbit around the sun. 
230 8
S on M
24 11
E on M
1.99 10 kg 3.84 10 m 2.18.
5.98 10 kg 1.50 10 m
F
F
⎛ ⎞⎛ ⎞× ×= =⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟× ×⎝ ⎠⎝ ⎠
 
EVALUATE: The force exerted by the sun is larger than the force exerted by the earth. The moon’s motion 
is a combination of orbiting the sun and orbiting the earth. 
 13.2. IDENTIFY: The gravity force between spherically symmetric spheres is 1 2g 2 ,
Gm mF
r
= where r is the 
separation between their centers. 
SET UP: 11 2 26.67 10 N m /kg .G −= × ⋅ The moment arm for the torque due to each force is 0.150 m. 
EXECUTE: (a) For each pair of spheres, 
11 2 2
7
g 2
(6.67 10 N m /kg )(1.10 kg)(25.0 kg) 1.27 10 N.
(0.120 m)
F
−
−× ⋅= = × 
From Figure 13.4 in the textbook we see that the forces for each pair are in opposite directions, so 
net 0.F = 
(b) The net torque is 7 8net g2 2(1.27 10 N)(0.150 m) 3.81 10 N m.F lτ
− −= = × = × ⋅ 
(c) The torque is very small and the apparatus must be very sensitive. The torque could be increased by 
increasing the mass of the spheres or by decreasing their separation. 
EVALUATE: The quartz fiber must twist through a measurable angle when a small torque is applied to it. 
 13.3. IDENTIFY: The gravitational attraction of the astronauts on each other causes them to accelerate toward 
each other, so Newton’s second law of motion applies to their motion. 
SET UP: The net force on each astronaut is the gravity force exerted by the other astronaut. Call the 
astronauts A and B, where 65 kgAm = and 72 kg.Bm = 
2
grav 1 2/F Gm m r= and .F maΣ = 
EXECUTE: (a) The free-body diagram for astronaut A is given in Figure 13.3a and for astronaut B in 
Figure 13.3b. 
 
GRAVITATION 
13
 
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Figure 13.3 
 
x xF maΣ = for A gives A A AF m a= and .AA
A
Fa
m
= And for B, .BB
B
Fa
m
= 
11 2 2 10
2 2
(65 kg)(72 kg)(6 673 10 N m /kg ) 7 807 10 N
(20 0 m)
A B
A B
m mF F G
r
− −= = = . × ⋅ = . ×
.
 so 
10
11 27 807 10 N 1 2 10 m/s
65 kgA
a
−
−. ×= = . × and 
10
11 27 807 10 N 1 1 10 m/s .
72 kgB
a
−
−. ×= = . × 
(b) Using constant-acceleration kinematics, we have 210 0 2 ,x xx x v t a t= + + which gives 
21
2A Ax a t= and 
21
2 .B Bx a t= 20 0 m,A Bx x+ = . so 
21
220 0 m ( )A Ba a t. = + and 
6
11 2 11 2
2(20 0 m) 1 32 10 s 15 days.
1 2 10 m/s 1 1 10 m/s
t − −
.= = . × =
. × + . ×
 
(c) Their accelerations would increase as they moved closer and the gravitational attraction between them 
increased. 
EVALUATE: Even though the gravitational attraction of the astronauts is much weaker than ordinary 
forces on earth, if it were the only force acting on the astronauts, it would produce noticeable effects. 
 13.4. IDENTIFY: Apply Eq. (13.2), generalized to any pair of spherically symmetric objects. 
SET UP: The separation of the centers of the spheres is 2R. 
EXECUTE: The magnitude of the gravitational attraction is 2 2 2 2/(2 ) /4 .GM R GM R= 
EVALUATE: Eq. (13.2) applies to any pair of spherically symmetric objects; one of the objects doesn’t 
have to be the earth. 
 13.5. IDENTIFY: Use Eq. (13.1) to find the force exerted by each large sphere. Add these forces as vectors to 
get the net force and then use Newton’s 2nd law to calculate the acceleration. 
SET UP: The forces are shown in Figure 13.5. 
 
 sin 0.80θ = 
cos 0.60θ = 
Take the origin of coordinate at point P. 
 
Figure 13.5 
 
EXECUTE: 112 2
(0.26 kg)(0.010 kg) 1.735 10 N
(0.100 m)
A
A
m mF G G
r
−= = = × 
11
2 1.735 10 N
B
B
m mF G
r
−= = × 
11 11sin (1.735 10 N)(0.80) 1.39 10 NAx AF F θ
− −= − = − × = − × 
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11 11cos (1.735 10 N)(0.60) 1.04 10 NAy AF F θ
− −= + = + × = + × 
11sin 1.39 10 NBx BF F θ
−= + = + × 
11cos 1.04 10 NBy BF F θ
−= + = + × 
x xF maΣ = gives Ax Bx xF F ma+ = 
0 xma= so 0xa = 
y yF maΣ = gives Ay By yF F ma+ = 
112(1.04 10 N) (0.010 kg) ya
−× = 
9 22.1 10 m/s ,ya
−= × directed downward midway between A and B 
EVALUATE: For ordinary size objects the gravitational force is very small, so the initial acceleration is 
very small. By symmetry there is no x-component of net force and the y-component is in the direction of 
the two large spheres, since they attract the small sphere. 
 13.6. IDENTIFY: The net force on A is the vector sum of the force due to B and the force due to C. In part (a), 
the two forces are in the same direction, but in (b) they are in opposite directions. 
SET UP: Use coordinates where x+ is to the right. Each gravitational force is attractive, so is toward the 
mass exerting it. Treat the masses as uniform spheres, so the gravitational force is the same as for point 
masses with the same center-to-center distances. The free-body diagrams for (a) and (b) are given in 
Figures 13.6a and 13.6b. The gravitational force is 2grav 1 2/ .F Gm m r= 
 
 
Figure 13.6 
 
EXECUTE: (a) Calling BF the force due to mass B and likewise for C, we have 
2
11 2 2 9
2 2
(2 00 kg)(6 673 10 N m /kg ) 1 069 10 N
(0 50 m)
A B
B
AB
m mF G
r
− −.= = . × ⋅ = . ×
.
 and 
2
11 2 2 8
2 2
(2 00 kg)(6 673 10 N m /kg ) 2 669 10 N.
(0 10 m)
A C
C
AC
m mF G
r
− −.= = . × ⋅ = . ×
.
 The net force is 
9 8 8
net, 1 069 10 N 2 669 10 N 2 8 10 Nx Bx CxF F F
− − −= + = . × + . × = . × to the right. 
(b) Following the same procedure as in (a), we have 
2
11 2 2 9
2 2
(2 00 kg)(6 673 10 N m /kg ) 1 668 10 N
(0 40 m)
A B
B
AB
m mF G
r
− −.= = . × ⋅ = . ×
.
 
2
11 2 2 8
2 2
(2 00 kg)(6 673 10 N m /kg ) 2 669 10 N
(0 10 m)
A C
C
AC
m mF G
r
− −.= = . × ⋅ = . ×
.
9 8 8
net, 1 668 10 N 2 669 10 N 2 5 10 Nx Bx CxF F F
− − −
 = + = . × − . × = − . × 
The net force on A is 82 5 10 N,−. × to the left. 
EVALUATE: As with any force, the gravitational force is a vector and must be treated like all other 
vectors. The formula 2grav 1 2/F Gm m r= only gives the magnitude of this force. 
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 13.7. IDENTIFY: The force exerted by the moon is the gravitational force, Mg 2 .
Gm mF
r
= The force exerted on 
the person by the earth is .w mg= 
SET UP: The mass of the moon is 22M 7.35 ×10 kg.m = 11 2 26.67 10 N m /kg .G −= × ⋅ 
EXECUTE: (a) 
22
11 2 2 3
moon g 8 2
(7.35 10 kg)(70 kg)(6.67 10 N m /kg ) 2.4 10 N.
(3.78 10 m)
F F − −×= = × ⋅ = ×
×
 
(b) 2earth (70 kg)(9.80 m/s ) 690 N.Fw= = = 
6
moon earth/ 3.5 10 .F F
−= × 
EVALUATE: The force exerted by the earth is much greater than the force exerted by the moon. The mass 
of the moon is less than the mass of the earth and the center of the earth is much closer to the person than is 
the center of the moon. 
 13.8. IDENTIFY: Use Eq. (13.2) to find the force each point mass exerts on the particle, find the net force, and 
use Newton’s second law to calculate the acceleration. 
SET UP: Each force is attractive. The particle (mass )m is a distance 1 0.200 mr = from 1 8.00 kgm = 
and therefore a distance 2 0.300 mr = from 2 15.0 kg.m = Let x+ be toward the 15.0 kg mass. 
EXECUTE: 11 2 2 811 2 2
1
(8.00 kg)(6.67 10 N m /kg ) (1.334 10 N/kg) ,
(0.200 m)
Gm m mF m
r
− −= = × ⋅ = × in the 
-direction.x− 11 2 2 822 2 2
2
(15.0 kg)(6.67 10 N m /kg ) (1.112 10 N/kg) ,
(0.300 m)
Gm m mF m
r
− −= = × ⋅ = × in the 
-direction.x+ The net force is 
8 8 9
1 2 ( 1.334 10 N/kg 1.112 10 N/kg) ( 2.2 10 N/kg) .x x xF F F m m
− − −= + = − × + × = − × 
9 22.2 10 m/s .xx
Fa
m
−= = − × The acceleration is 9 22.2 10 m/s ,−× toward the 8.00 kg mass. 
EVALUATE: The smaller mass exerts the greater force, because the particle is closer to the smaller mass. 
 13.9. IDENTIFY: Use Eq. (13.2) to calculate the gravitational force each particle exerts on the third mass. The 
equilibrium is stable when for a displacement from equilibrium the net force is directed toward the 
equilibrium position and it is unstable when the net force is directed away from the equilibrium position. 
SET UP: For the net force to be zero, the two forces on M must be in opposite directions. This is the case 
only when M is on the line connecting the two particles and between them. The free-body diagram for M 
 is given in Figure 13.9. 1 3m m= and 2 .m m= If M is a distance x from 1,m it is a distance 1.00 m x− 
from 2.m 
EXECUTE: (a) 2 21 2 2 2
3
0. 3 (1.00 m ) .
(1.00 m )x x x
mM mM
F F F G G x x
x x
= + = − + = − =
−
 
1.00 m / 3.x x− = ± Since M is between the two particles, x must be less than 1.00 m and 
1.00 m 0.634 m.
1 1/ 3
x = =
+
 M must be placed at a point that is 0.634 m from the particle of mass 3m and 
0.366 m from the particle of mass m. 
(b) (i) If M is displaced slightly to the right in Figure 13.9, the attractive force from m is larger than the 
force from 3m and the net force is to the right. If M is displaced slightly to the left in Figure 13.9, the 
attractive force from 3m is larger than the force from m and the net force is to the left. In each case the 
net force is away from equilibrium and the equilibrium is unstable. 
(ii) If M is displaced a very small distance along the y axis in Figure 13.9, the net force is directed opposite 
to the direction of the displacement and therefore the equilibrium is stable. 
EVALUATE: The point where the net force on M is zero is closer to the smaller mass. 
 
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Figure 13.9 
 
 13.10. IDENTIFY: The force 1F exerted by m on M and the force 2F exerted by 2m on M are each given by 
Eq. (13.2) and the net force is the vector sum of these two forces. 
SET UP: Each force is attractive. The forces on M in each region are sketched in Figure 13.10a. Let M be 
at coordinate x on the x-axis. 
EXECUTE: (a) For the net force to be zero, 1F and 2F must be in opposite directions and this is the case 
only for 0 .x L< < 1 2 0+F F = then requires 1 2.F F= 2 2
(2 ) .
( )
GmM G m M
x L x
=
−
 2 22 ( )x L x= − and 
2 .L x x x− = ± must be less than L, so 0.414 .
1 2
Lx L= =
+
 
(b) For 0,x < 0.xF > 0xF → as x → −∞ and xF → +∞ as 0.x → For ,x L> 0.xF < 0xF → as 
x → ∞ and xF → −∞ as .x L→ For 0 0.414 ,x L< < 0xF < and xF increases from −∞ to 0 as x goes 
from 0 to 0.414L. For 0.414 ,L x L< < 0xF > and xF increases from 0 to +∞ as x goes from 0.414L to L. 
The graph of xF versus x is sketched in Figure 13.10b. 
EVALUATE: Any real object is not exactly a point so it is not possible to have both m and M exactly at 
0x = or 2m and M both exactly at .x L= But the magnitude of the gravitational force between two objects 
approaches infinity as the objects get very close together. 
 
 
Figure 13.10 
 
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 13.11. IDENTIFY: Eg 2 ,
mmF G
r
= so Eg 2 ,
ma G
r
= where r is the distance of the object from the center of the earth. 
SET UP: E+ ,r h R= where h is the distance of the object above the surface of the earth and 
6
E 6.38 10 mR = × is the radius of the earth. 
EXECUTE: To decrease the acceleration due to gravity by one-tenth, the distance from the center of the 
earth must be increased by a factor of 10, and so the distance above the surface of the earth is 
7
E( 10 1) 1.38 10 m.R− = × 
EVALUATE: This height is about twice the radius of the earth. 
 13.12. IDENTIFY: Apply Eq. (13.4) to the earth and to Venus. .w mg= 
SET UP: 2E2
E
9.80 m/s .Gmg
R
= = V E0.815m m= and V E0.949 .R R= E E 75.0 N.w mg= = 
EXECUTE: (a) V E EV E2 2 2
V E E
(0.815 ) 0.905 0.905 .
(0.949 )
Gm G m Gmg g
R R R
= = = = 
(b) V V E0.905 (0.905)(75.0 N) 67.9 N.w mg mg= = = = 
EVALUATE: The mass of the rock is independent of its location but its weight equals the gravitational 
force on it and that depends on its location. 
 13.13. (a) IDENTIFY and SET UP: Apply Eq. (13.4) to the earth and to Titania. The acceleration due to gravity at 
the surface of Titania is given by 2T T T/ ,g Gm R= where Tm is its mass and TR is its radius. 
For the earth, 2E E E/ .g Gm R= 
EXECUTE: For Titania, T E/1700m m= and T E/8,R R= so 
T E E
T E2 2 2
T E E
( /1700) 64 0.0377 .
1700( /8)
Gm G m Gmg g
R R R
⎛ ⎞= = = =⎜ ⎟
⎝ ⎠
 
Since 2E 9.80 m/s ,g = 
2 2
T (0.0377)(9.80 m/s ) 0.37 m/s .g = = 
EVALUATE: g on Titania is much smaller than on earth. The smaller mass reduces g and is a greater effect 
than the smaller radius, which increases g. 
(b) IDENTIFY and SET UP: Use density mass/volume.= Assume Titania is a sphere. 
EXECUTE: From Section 13.2 we know that the average density of the earth is 35500 kg/m . For Titania 
3 3T
T E3 34 4
T E3 3
/1700 512 512 (5500 kg/m ) 1700 kg/m .
1700 1700( /8)
Em m
R R
ρ ρ
π π
= = = = = 
EVALUATE: The average density of Titania is about a factor of 3 smaller than for earth. We can write 
Eq. (13.4) for Titania as 4T T T3 .g GRπ ρ= T Eg g< both because T Eρ ρ< and T E.R R< 
 13.14. IDENTIFY: Apply Eq. (13.4) to Rhea. 
SET UP: / .m Vρ = The volume of a sphere is 343 .V Rπ= 
EXECUTE: 
2
212.44 10 kggRM G= = × and 
3 3
3 1.30 10 kg/m .(4 /3)
M
R
ρ
π
= = × 
EVALUATE: The average density of Rhea is about one-fourth that of the earth. 
 13.15. IDENTIFY: Apply Eq. (13.2) to the astronaut. 
SET UP: 24E 5.97 10 kgm = × and 
6
E 6.38 10 m.R = × 
EXECUTE: Eg 2 .
mmF G
r
= 3 E600 10 mr R= × + so g 610 N.F = At the surface of the earth, 
g 735 N.w m= = The gravity force is not zero in orbit. The satellite and the astronaut have the same 
acceleration so the astronaut’s apparent weight is zero. 
EVALUATE: In Eq. (13.2), r is the distance of the object from the center of the earth. 
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 13.16. IDENTIFY: The gravity of Io limits the height to which volcanic material will rise. The acceleration due to 
gravity at the surface of Io depends on its mass and radius. 
SET UP: The radius of Io is 61 815 10 m.R = . × Use coordinates where y+ is upward. At the maximum 
height, 0 0,yv = Io ,ya g= − which is assumed to be constant. Therefore the constant-acceleration 
kinematics formulas apply. The acceleration due to gravity at Io’s surface is given by 2Io / .g Gm R= 
SOLVE: At the surface of Io, 
11 2 2 22
2
Io 2 6 2
(6 673 10 N m /kg )(8 94 10 kg) 1 81 m/s .
(1 815 10 m)
Gmg
R
−. × ⋅ . ×= = = .
. ×
 For 
constant acceleration (assumed), the equation 2 20 02 ( )y y yv v a y y= + − applies, so 
2 5 3
0 02 ( ) 2( 1 81 m/s )(5 00 10 m) 1 345 10 m/s.y yv a y y= − − = − − . . × = . × Now solve for 0y y− when 
3
0 1 345 10 m/syv = . × and 
29 80 m/s .ya = − . The equation 
22
00 2 ( )yy yv v a y y= + − gives 
22 3 2
0
0 2
(1 345 10 m/s) 92 km.
2 2( 9 80 m/s )
yy
y
v v
y y
a
− − . ×− = = =
− .
 
EVALUATE: Even though the mass of Io is around 100 times smaller than that of the earth, the 
acceleration due to gravity at its surface is only about 1/6 of that of the earth because Io’s radius is much 
smaller than earth’s radius. 
 13.17. IDENTIFY: The escape speed, from the results of Example 13.5, is 2 / .GM R 
SET UP: For Mars, 236.42 10 kgM = × and 63.40 10 m.R = × For Jupiter, 271.90 10 kgM = × and 
76.91 10 m.R = × 
EXECUTE: (a) 11 2 2 23 6 32(6.673 10 N m /kg )(6.42 10 kg)/(3.40 10 m) 5.02 10 m/s.v −= × ⋅ × × = × 
(b) 11 2 2 27 7 42(6.673 10 N m /kg (1.90 10 kg)/(6.91 10 m) 6.06 10 m/s.v −= × ⋅ × × = × 
(c) Both the kinetic energy and the gravitational potential energy are proportional to the mass of the 
spacecraft. 
EVALUATE: Example 13.5 calculates the escape speed for earth to be 41.12 10 m/s.× This is larger than 
our result for Mars and less than our result for Jupiter. 
 13.18. IDENTIFY: The kinetic energy is 212K mv= and the potential energy is .
GMm
U
r
= − 
SET UP: The mass of the earth is 24E 5.97 10 kg.M = × 
EXECUTE: (a) 3 2 912 (629 kg)(3.33 10 m/s) 3.49 10 JK = × = × 
(b) 
11 2 2 24
7E
9
(6.673 10 N m /kg )(5.97 10 kg)(629 kg) 8.73 10 J.
2.87 10 m
GM mU
r
−× ⋅ ×= − = − = − ×
×
 
EVALUATE: The total energy K U+ is positive. 
 13.19. IDENTIFY: Apply Newton’s second law to the motion of the satellite and obtain an equation that relates 
the orbital speed v to the orbital radius r. 
SET UP: The distances are shown in Figure 13.19a. 
 
 The radius of the orbit is E .r h R= + 
5 6 67.80 10 m 6.38 10 m 7.16 10 m.r = × + × = × 
Figure 13.19a 
 
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The free-body diagram for the satellite is given in Figure 13.19b. 
 
 (a) EXECUTE: y yF maΣ = 
g radF ma= 
2
2
Emm vG m
rr
= 
Figure 13.19b 
 
11 2 2 24
3E
6
(6.673 10 N m /kg )(5.97 10 kg) 7.46 10 m/s
7.16 10 m
Gmv
r
−× ⋅ ×= = = ×
×
 
(b) 
6
3
2 2 (7.16 10 m) 6030 s 1.68 h.
7.46 10 m/s
rT
v
π π ×= = = =
×
 
EVALUATE: Note that Er h R= + is the radius of the orbit, measured from the center of the earth. For this 
satellite r is greater than for the satellite in Example 13.6, so its orbital speed is less. 
 13.20. IDENTIFY: The time to complete one orbit is the period T, given by Eq. (13.12). The speed v of the 
satellite is given by 2 .rv
T
π= 
SET UP: If h is the height of the orbit above the earth’s surface, the radius of the orbit is E.r h R= + 
6
E 6.38 10 mR = × and 
24
E 5.97 10 kg.m = × 
EXECUTE: (a) 
3/2 5 6 3/2
3
11 2 2 24E
2 2 (7.05 10 m 6.38 10 m) 5.94 10 s 99.0 min
(6.67 10 N m /kg )(5.97 10 kg)
rT
Gm
π π
−
× + ×= = = × =
× ⋅ ×
 
(b) 
5 6
3
3
2 (7.05 10 m 6.38 10 m) 7.49 10 m/s 7.49 km/s
5.94 10 s
v π × + ×= = × =
×
 
EVALUATE: The satellite in Example 13.6 is at a lower altitude and therefore has a smaller orbit radius 
than the satellite in this problem. Therefore, the satellite in this problem has a larger period and a smaller 
orbital speed. But a large percentage change in h corresponds to a small percentage change in r and the 
values of T and v for the two satellites do not differ very much. 
 13.21. IDENTIFY: We know orbital data (speed and orbital radius) for one satellite and want to use it to find the 
orbital speed of another satellite having a known orbital radius. Newton’s second law and the law of 
universal gravitation apply to both satellites. 
SET UP: For circular motion, 2net / ,F ma mv r= = which in this case is 
2
p
2 .
mm vG m
rr
= 
EXECUTE: Using 
2
p
2 ,
mm vG m
rr
= we get 2p constant.Gm rv= = 
2 2
1 1 2 2 .r v r v= 
7
1
2 1 7
2
5 00 10 m(4800 m/s) 6200 m/s.
3 00 10 m
rv v
r
. ×= = =
. ×
 
EVALUATE: The more distant satellite moves slower than the closer satellite, which is reasonable since 
the planet’s gravity decreases with distance. The masses of the satellites do not affect their orbits. 
 13.22. IDENTIFY: We can calculate the orbital period T from the number of revolutions per day. Then the period 
and the orbit radius are related by Eq. (13.12). 
SET UP: 24E 5.97 10 kgm = × and 6E 6.38 10 m.R = × The height h of the orbit above the surface of the 
earth is related to the orbit radius r by E.r h R= + 
41day 8.64 10 s.= × 
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EXECUTE: The satellite moves 15.65 revolutions in 48.64 10 s,× so the time for 1.00 revolution is 
4
38.64 10 s 5.52 10 s.
15.65
T ×= = × 
3/2
E
2 rT
Gm
π= gives 
1/3 1/32 11 2 2 24 3 2
E
2 2
[6.67 10 N m /kg ][5.97 10 kg][5.52 10 s] .
4 4
Gm Tr
π π
−⎛ ⎞ ⎛ ⎞× ⋅ × ×= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
 66.75 10 mr = × and 
5
E 3.7 10 m 370 km.h r R= − = × = 
EVALUATE: The period of this satellite is slightly larger than the period for the satellite in Example 13.6 
and the altitude of this satellite is therefore somewhat greater. 
 13.23. IDENTIFY: Apply mΣ =F a to the motion of the baseball. 2 .rv
T
π= 
SET UP: 3D 6 10 m.r = × 
EXECUTE: (a) g radF ma= gives 
2
D
2
DD
.m m vG m
rr
= 
11 2 2 15
D
3
D
(6.673 10 N m /kg )(2.0 10 kg) 4.7 m/s
6 10 m
Gmv
r
−× ⋅ ×= = =
×
 
4.7 m/s 11 mph,= which is easy to achieve. 
(b) 
32 2 (6 10 m)
8020 s 134 min 2.23 h.
4.7 m/s
r
T
v
π π ×
= = = = = The game would last a long time. 
EVALUATE: The speed v is relative to the center of Deimos. The baseball would already have some speed 
before we throw it, because of the rotational motion of Deimos. 
 13.24. IDENTIFY: 2 rT
v
π= and g rad.F ma= 
SET UP: The sun has mass 30S 1.99 10 kg.m = × The radius of Mercury’s orbit is 105.79 10 m,× so the 
radius of Vulcan’s orbit is 103.86 10 m.× 
EXECUTE: g radF ma= gives 
2
S
2
m m vG m
rr
= and 2 S .Gmv
r
= 
3/2 10 3/2
6
11 2 2 30S S
2 2 (3.86 10 m)2 4.13 10 s 47.8 days
(6.673 10 N m /kg )(1.99 10 kg)
r rT r
Gm Gm
π ππ
−
×= = = = × =
× ⋅ ×
 
EVALUATE: The orbital period of Mercury is 88.0 d, so we could calculate T for Vulcan as 
3/2(88.0 d)(2/3) 47.9 days.T = = 
 13.25. IDENTIFY: The orbital speed is given by / ,v Gm r= where m is the mass of the star.The orbital period is 
given by 2 .rT
v
π= 
SET UP: The sun has mass 30S 1.99 10 kg.m = × The orbit radius of the earth is 
111.50 10 m.× 
EXECUTE: (a) / .v Gm r= 
11 2 2 30 11 4(6.673 10 N m /kg )(0.85 1.99 10 kg)/((1.50 10 m)(0.11)) 8.27 10 m/s.v −= × ⋅ × × × = × 
(b) 62 / 1.25 10 s 14.5 daysr vπ = × = (about two weeks). 
EVALUATE: The orbital period is less than the 88-day orbital period of Mercury; this planet is orbiting 
very close to its star, compared to the orbital radius of Mercury. 
 13.26. IDENTIFY: The period of each satellite is given by Eq. (13.12). Set up a ratio involving T and r. 
SET UP: 
3/2
p
2 rT
Gm
π= gives 3/2
p
2 constant,T
Gmr
π= = so 1 23/2 3/2
1 2
.T T
r r
= 
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EXECUTE: 
3/2 3/2
2
2 1
1
48,000 km(6.39 days) 24.5 days.
19,600 km
rT T
r
⎛ ⎞ ⎛ ⎞
= = =⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
 For the other satellite, 
3/2
2
64,000 km(6.39 days) 37.7 days.
19,600 km
T
⎛ ⎞
= =⎜ ⎟
⎝ ⎠
 
EVALUATE: T increases when r increases. 
 13.27. IDENTIFY: In part (b) apply the results from part (a). 
SET UP: For Pluto, 0.248e = and 125.92 10 m.a = × For Neptune, 0.010e = and 124.50 10 m.a = × The 
orbital period for Pluto is 247.9 y.T = 
EXECUTE: (a) The result follows directly from Figure 13.18 in the textbook. 
(b) The closest distance for Pluto is 12 12(1 0.248)(5.92 10 m) 4.45 10 m.− × = × The greatest distance for 
Neptune is 12 12(1 0.010)(4.50 10 m) 4.55 10 m.+ × = × 
(c) The time is the orbital period of Pluto, 248 y.T = 
EVALUATE: Pluto’s closest distance calculated in part (a) is 12 80.10 10 m 1.0 10 km,× = × so Pluto is 
about 100 million km closer to the sun than Neptune, as is stated in the problem. The eccentricity of 
Neptune’s orbit is small, so its distance from the sun is approximately constant. 
 13.28. IDENTIFY: 
3/2
star
2 ,rT
Gm
π= where starm is the mass of the star. 
2 .rv
T
π= 
SET UP: 53.09 days 2.67 10 s.= × The orbit radius of Mercury is 105.79 10 m.× The mass of our sun is 
301.99 10 kg.× 
EXECUTE: (a) 52.67 10 s.T = × 10 9(5.79 10 m)/9 6.43 10 m.r = × = × 
3/2
star
2 rT
Gm
π= gives 
2 3 2 9 3
30
star 2 5 2 11 2 2
4 4 (6.43 10 m) 2.21 10 kg
(2.67 10 s) (6.67 10 N m /kg )
rm
T G
π π
−
×= = = ×
× × ⋅
. star
sun
1.11,m
m
= so 
star sun1.11 .m m= 
(b) 
9
5
5
2 2 (6.43 10 m)
1.51 10 m/s 151 km/s
2.67 10 s
r
v
T
π π ×
= = = × =
×
 
EVALUATE: The orbital period of Mercury is 88.0 d. The period for this planet is much less primarily 
because the orbit radius is much less and also because the mass of the star is greater than the mass of our 
sun. 
 13.29. IDENTIFY: Knowing the orbital radius and orbital period of a satellite, we can calculate the mass of the 
object about which it is revolving. 
SET UP: The radius of the orbit is 910 5 10 mr = . × and its period is 56.3 days 5.443 10 s.T = = × The 
mass of the sun is 30S 1 99 10 kg.m = . × The orbital period is given by 
3/2
HD
2 .rT
Gm
π= 
EXECUTE: Solving 
3/2
HD
2 rT
Gm
π= for the mass of the star gives 
2 3 2 9 3
30
HD 2 5 2 11 2 2
4 4 (10 5 10 m) 2 3 10 kg,
(5 443 10 s) (6 673 10 N m /kg )
rm
T G
π π
−
. ×= = = . ×
. × . × ⋅
 which is HD S1 2 .m m= . 
EVALUATE: The mass of the star is only 20% greater than that of our sun, yet the orbital period of the 
planet is much shorter than that of the earth, so the planet must be much closer to the star than the earth is. 
 13.30. IDENTIFY: Section 13.6 states that for a point mass outside a spherical shell the gravitational force is the 
same as if all the mass of the shell were concentrated at its center. It also states that for a point inside a 
spherical shell the force is zero. 
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